I've been trying to figure out how I could evaluate the following infinite product.
I am apparently supposed to find the exact value, but I don't see any way I could do this.
$$\prod_{n=0}^\infty \frac{{(4n+2)}^{\frac{3}{4n+2}}}{{(4n+3)}^{\frac{1}{4n+3}}{(4n+4)}^{\frac{1}{4n+4}}{(4n+5)}^{\frac{1}{4n+5}}}$$
I don't wish to know the solution I just want to know what techniques I should learn to be able to find the solution.
(edit) I forgot to write that the top of the fraction is cubed. Of course this problem would be simple if the power was 1.
The original version of your infinite product can be written in the following form: $$ \exp\sum_{n\geq 0}\underbrace{\left[\frac{\log(4n+2)}{4n+2}-\frac{\log(4n+3)}{4n+3}-\frac{\log(4n+4)}{4n+4}-\frac{\log(4n+5)}{4n+5}\right]}_{\approx -2\frac{\log(4n+4)}{4n+4}}$$ the series is quite blatantly divergent to $-\infty$, hence the associated product is convergent to zero.
Interesting things happen if we change a sign: $$ \exp\sum_{n\geq 0}\underbrace{\left[\frac{\log(4n+2)}{4n+2}-\frac{\log(4n+3)}{4n+3}+\frac{\log(4n+4)}{4n+4}-\frac{\log(4n+5)}{4n+5}\right]}_{\approx 2\frac{\log(4n+4)}{(4n+4)^2}}$$ equals ${\large\prod_{n\geq 0}}\frac{(4n+2)^{\frac{1}{4n+2}}(4n+4)^{\frac{1}{4n+4}}}{(4n+3)^{\frac{1}{4n+3}}(4n+5)^{\frac{1}{4n+5}}}$ or
$$\exp\sum_{n\geq 1}\frac{(-1)^{n}\log n}{n}=\exp\eta'(1)=\exp\left[\gamma\log 2-\frac{1}{2}\log^2 2\right]=2^{\gamma-\frac{\log 2}{2}}. $$
Similarly, $$ \exp\sum_{n\geq 0}\underbrace{\left[3\cdot\frac{\log(4n+2)}{4n+2}-\frac{\log(4n+3)}{4n+3}-\frac{\log(4n+4)}{4n+4}-\frac{\log(4n+5)}{4n+5}\right]}_{\approx 6\frac{\log(4n+4)}{(4n+4)^2}}$$ can be written as $$\exp\underbrace{\left[\sum_{n\geq 1}\frac{(-1)^n \log n}{n}+2\sum_{n\geq 1}\frac{(-1)^{n+1} \log(2n)}{2n}\right]}_{\text{massive cancellation!}}=\color{red}{\exp\log^2(2)}=2^{\log 2}. $$